In triangle ABC you have:
BC^2 = AB^2 + AC^2 - 2AB.AC.cos A
BC^2=AB^2+AC^2(you may use this because Pyth sais that this is true).
Out the last 2 lines follows: 2.AB.AC. cos A
Ab <> 0 and AC <> 0
so: cos A = 0 and A = 90 degrees
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Second proof
Let''s say that A<90 degrees. In triangle ABC you have the line from C which stands perpendicular on AB(don''t now what the english word is, in dutch we call it: "Hoogtelijn").
In triangle ABC: AD^2+DC^2=AC^2
In triangle DBC: DB^2+DC^2=BC^2
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AD^2-DB^2=AC^2-BC^2
BC^2=AB^2+AC^2
so: AD^2-DB^2=AC^2-(AB^2+AC^2)
AD^2-DB^2=-AB^2
AD^2+AB^2=DB^2
And that is not possible because DB would be bigger then AB. So if the angle is less then 90 degrees pythagoras is not true.
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Lets say A > 90 degrees
You have triangle ABC and DBC
AD^2+AB^2=DB^2=(AD+AB)^2
AD^2+AB^2=AD^2+2.AD.AB+AB^2
2.AD.AB=0
But AD and AB <> 0 so this is not possible.
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If A is less then 90 degrees it failes, and when it''s more then 90 it failes also. So it needs to be 90
Woooh, what a story :D I''m glad that I still have my mathbook laying around
